Stainless steel, tell us about its properties and what should be taken into account when using it?

What is the maximum resultant intensity in terms of the intensity of the initial unpolarized incoming light io?

Dennis_Churaev [7]

The unpolarized light becomes less intense. When unpolarized light is run through a polarizer, its intensity is diminished by a factor of ½.

<h3>Unpolarized light:</h3>

Unpolarized light is diminished by a factor of ½ when it travels through a polarizer. Cos 2θ averages out across all angles to be ½ in size. I transmitted = I₀ and ½ I₀ at all angles The unpolarized light's intensity decreases to zero.

The light-producing atoms on the heated filament's surface behave independently of one another. Each of these emissions can be roughly represented as a brief "wave train" lasting between approximately 109 and 108 seconds. These wave trains, each with its unique polarization direction, are superimposed to form the electromagnetic wave that is coming from the filament.

Unpolarized light is produced by all common light sources, including the Sun, incandescent and fluorescent lighting, and flames.

Learn more about unpolarized light here:

brainly.com/question/15562548

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6 0

1 year ago

Cowbirds follow groups of cows, or other grazing animals, eating the insects kicked up by the walking animals. The herbivores ar

Jobisdone [24]

Answer:

d: mutualism

Explanation:

I'm guessing that means they're on mutual grounds so they don't mind sharing?

8 0

3 years ago

A snowboarder starts from rest on an inclined planar snow surface and gives herself a tiny push to start sliding down the hill.

tiny-mole [99]

Answer:

v = 26.52 m/s

Explanation:

Here we know that as she passes the flag she will gain some speed

so here we can say that

Work done by all forces = change in kinetic energy of the system

so we will have

$W_g + W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

here we know that

$W_g = mgh$

$W_f = -\mu mg d$

$v_i = 0$

h = 39.78 m

d = 27.81 m

so we have

$mgh - \mu mg d = \frac{1}{2}mv^2 - 0$

$(9.8)(39.78 - (0.14)(27.81)) = \frac{1}{2} v^2$

$v = 26.52 m/s$

8 0

3 years ago

5. A 5.5 x10-6 C charge is located 0.28 m from a -3.5 x 10-6 C charge.

ra1l [238]

(a) The magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

Explanation:

The force acting between two charged particles separated by a distance is termed as Coloumb's force or electrostatic force. It can be termed as electrostatic force of attraction if the the force acting between the charges are oppositely charged. And it can be termed as electrostatic force of repulsion if the charges are similar or like charges.

In the present case, there is a positive and negative charge, so electrostatic force of attraction will be acting between them. As per Coloumb's law, the electrostatic force of attraction is directly proportional to the product of charges and inversely proportional to the square of distance of separation.

$F = \frac{kQq}{d^{2} }$

Here, k is the constant of proportionality which is equal to 9 × $10^{9}$ and Q, q are the two charges, d is the distance of separation.

So here Q = 5.5 × $10^{-6} C$ and q = - 3.5 × $10^{-6} C$ and d = 0.28 m

Then, $F=-\frac{9*10^{9}*5.5*10^{-6} * 3.5 * 10^{-6} }{(0.28)^{2} } = 2209.82*10^{-3}$

So the magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

3 0

4 years ago

You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo

IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

= (0.81) (1.42) (μ m g)

= (0.81) (1.42) (200)

= 230 N

4 0

3 years ago